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CHAPTER 4
REACTIONS IN AQUEOUS SOLUTIONS
4.7
4.8
4.9
(a) is a strong electrolyte. The compound dissociates completely into ions in solution.
(b) is a nonelectrolyte. The compound dissolves in water, but the molecules remain intact.
(c) is a weak electrolyte. A small amount of the compound dissociates into ions in water.
When NaCl dissolves in water it dissociates into Na and Cl ions. When the ions are hydrated, the water
molecules will be oriented so that the negative end of the water dipole interacts with the positive sodium ion,
and the positive end of the water dipole interacts with the negative chloride ion. The negative end of the
water dipole is near the oxygen atom, and the positive end of the water dipole is near the hydrogen atoms.
The diagram that best represents the hydration of NaCl when dissolved in water is choice (c).
Ionic compounds, strong acids, and strong bases (metal hydroxides) are strong electrolytes (completely
broken up into ions of the compound). Weak acids and weak bases are weak electrolytes. Molecular
substances other than acids or bases are nonelectrolytes.
(a) very weak electrolyte
(b)
strong electrolyte (ionic compound)
(c) strong electrolyte (strong acid)
(d)
weak electrolyte (weak acid)
(e) nonelectrolyte (molecular compound - neither acid nor base)
4.10
Ionic compounds, strong acids, and strong bases (metal hydroxides) are strong electrolytes (completely
broken up into ions of the compound). Weak acids and weak bases are weak electrolytes. Molecular
substances other than acids or bases are nonelectrolytes.
(a) strong electrolyte (ionic)
(b)
nonelectrolyte
(c) weak electrolyte (weak base)
(d)
strong electrolyte (strong base)
4.11
Since solutions must be electrically neutral, any flow of positive species (cations) must be balanced by the
flow of negative species (anions). Therefore, the correct answer is (d).
4.12
(a) Solid NaCl does not conduct. The ions are locked in a rigid lattice structure.
(b) Molten NaCl conducts. The ions can move around in the liquid state.
(c) Aqueous NaCl conducts. NaCl dissociates completely to Na(aq) and Cl(aq) in water.
4.13
4.14
Measure the conductance to see if the solution carries an electrical current. If the solution is conducting, then
you can determine whether the solution is a strong or weak electrolyte by comparing its conductance with that
of a known strong electrolyte.
Since HCl dissolved in water conducts electricity, then HCl(aq) must actually exists as H(aq) cations and
Cl(aq) anions. Since HCl dissolved in benzene solvent does not conduct electricity, then we must assume
that the HCl molecules in benzene solvent do not ionize, but rather exist as un-ionized molecules.
70
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.17
Refer to Table 4.2 of the text to solve this problem. AgCl is insoluble in water. It will precipitate from
solution. NaNO3 is soluble in water and will remain as Na and NO3 ions in solution. Diagram (c) best
represents the mixture.
4.18
Refer to Table 4.2 of the text to solve this problem. Mg(OH) 2 is insoluble in water. It will precipitate from
solution. KCl is soluble in water and will remain as K and Cl ions in solution. Diagram (b) best represents
the mixture.
4.19
Refer to Table 4.2 of the text to solve this problem.
(a)
(b)
(c)
(d)
4.20
Ca3(PO4)2 is insoluble.
Mn(OH)2 is insoluble.
AgClO3 is soluble.
K2S is soluble.
Strategy: Although it is not necessary to memorize the solubilities of compounds, you should keep in mind
the following useful rules: all ionic compounds containing alkali metal cations, the ammonium ion, and the
nitrate, bicarbonate, and chlorate ions are soluble. For other compounds, refer to Table 4.2 of the text.
Solution:
(a) CaCO3 is insoluble. Most carbonate compounds are insoluble.
(b) ZnSO4 is soluble. Most sulfate compounds are soluble.
(c) Hg(NO3)2 is soluble. All nitrate compounds are soluble.
(d) HgSO4 is insoluble. Most sulfate compounds are soluble, but those containing Ag, Ca2, Ba2, Hg2,
and Pb2 are insoluble.
(e) NH4ClO4 is soluble. All ammonium compounds are soluble.
4.21
(a)
Ionic: 2Ag(aq)  2NO3(aq)  2Na(aq)  SO42(aq) 
 Ag2SO4(s)  2Na(aq)  2NO3(aq)
Net ionic: 2Ag(aq)  SO42(aq) 
 Ag2SO4(s)
(b)
Ionic: Ba2(aq)  2Cl(aq)  Zn2(aq)  SO42(aq) 
 BaSO4(s)  Zn2(aq)  2Cl(aq)
Net ionic: Ba2(aq)  SO42(aq) 
 BaSO4(s)
(c)
Ionic: 2NH4(aq)  CO32(aq)  Ca2(aq)  2Cl(aq) 
 CaCO3(s)  2NH4(aq)  2Cl(aq)
Net ionic: Ca2(aq)  CO32(aq) 
 CaCO3(s)
4.22
(a)
Strategy: Recall that an ionic equation shows dissolved ionic compounds in terms of their free ions. A net
ionic equation shows only the species that actually take part in the reaction. What happens when ionic
compounds dissolve in water? What ions are formed from the dissociation of Na 2S and ZnCl2? What
happens when the cations encounter the anions in solution?
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
71
Solution: In solution, Na2S dissociates into Na and S2 ions and ZnCl2 dissociates into Zn2 and Cl ions.
According to Table 4.2 of the text, zinc ions (Zn2) and sulfide ions (S2) will form an insoluble compound,
zinc sulfide (ZnS), while the other product, NaCl, is soluble and remains in solution. This is a precipitation
reaction. The balanced molecular equation is:
Na2S(aq)  ZnCl2(aq) 
 ZnS(s)  2NaCl(aq)
The ionic and net ionic equations are:
Ionic: 2Na(aq)  S2(aq)  Zn2(aq)  2Cl(aq) 
 ZnS(s)  2Na(aq)  2Cl(aq)
Net ionic: Zn2(aq)  S2(aq) 
 ZnS(s)
Check: Note that because we balanced the molecular equation first, the net ionic equation is balanced as to
the number of atoms on each side, and the number of positive and negative charges on the left-hand side of
the equation is the same.
(b)
Strategy: What happens when ionic compounds dissolve in water? What ions are formed from the
dissociation of K3PO4 and Sr(NO3)2? What happens when the cations encounter the anions in solution?
Solution: In solution, K3PO4 dissociates into K and PO43 ions and Sr(NO3)2 dissociates into Sr2 and
NO3 ions. According to Table 4.2 of the text, strontium ions (Sr 2) and phosphate ions (PO43) will form an
insoluble compound, strontium phosphate [Sr3(PO4)2], while the other product, KNO3, is soluble and remains
in solution. This is a precipitation reaction. The balanced molecular equation is:
2K3PO4(aq)  3Sr(NO3)2(aq) 
 Sr3(PO4)2(s)  6KNO3(aq)
The ionic and net ionic equations are:
Ionic: 6K(aq)  2PO43(aq)  3Sr2(aq)  6NO3(aq) 
 Sr3(PO4)2(s)  6K(aq)  6NO3(aq)
Net ionic: 3Sr2(aq)  2PO43(aq) 
 Sr3(PO4)2(s)
Check: Note that because we balanced the molecular equation first, the net ionic equation is balanced as to
the number of atoms on each side, and the number of positive and negative charges on the left-hand side of
the equation is the same.
(c)
Strategy: What happens when ionic compounds dissolve in water? What ions are formed from the
dissociation of Mg(NO3)2 and NaOH? What happens when the cations encounter the anions in solution?
Solution: In solution, Mg(NO3)2 dissociates into Mg2 and NO3 ions and NaOH dissociates into Na and
OH ions. According to Table 4.2 of the text, magnesium ions (Mg2) and hydroxide ions (OH) will form an
insoluble compound, magnesium hydroxide [Mg(OH)2], while the other product, NaNO3, is soluble and
remains in solution. This is a precipitation reaction. The balanced molecular equation is:
Mg(NO3)2(aq)  2NaOH(aq) 
 Mg(OH)2(s)  2NaNO3(aq)
The ionic and net ionic equations are:
Ionic: Mg2(aq)  2NO3(aq)  2Na(aq)  2OH(aq) 
 Mg(OH)2(s)  2Na(aq)  2NO3(aq)
Net ionic: Mg2(aq)  2OH(aq) 
 Mg(OH)2(s)
72
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
Check: Note that because we balanced the molecular equation first, the net ionic equation is balanced as to
the number of atoms on each side, and the number of positive and negative charges on the left-hand side of
the equation is the same.
4.23
(a)
Both reactants are soluble ionic compounds. The other possible ion combinations, Na 2SO4 and
Cu(NO3)2, are also soluble.
(b)
Both reactants are soluble. Of the other two possible ion combinations, KCl is soluble, but BaSO 4 is
insoluble and will precipitate.
Ba2(aq)  SO42(aq)  BaSO4(s)
4.24
4.31
(a)
Add chloride ions. KCl is soluble, but AgCl is not.
(b)
Add hydroxide ions. Ba(OH)2 is soluble, but Pb(OH)2 is insoluble.
(c)
Add carbonate ions. (NH4)2CO3 is soluble, but CaCO3 is insoluble.
(d)
Add sulfate ions. CuSO4 is soluble, but BaSO4 is insoluble.
(a)
HI dissolves in water to produce H and I, so HI is a Brønsted acid.
(b)
CH3COO can accept a proton to become acetic acid CH3COOH, so it is a Brønsted base.
(c)
(d)
4.32
H2PO4 can either accept a proton, H, to become H3PO4 and thus behaves as a Brønsted base, or can
donate a proton in water to yield H and HPO42, thus behaving as a Brønsted acid.
HSO4 can either accept a proton, H, to become H2SO4 and thus behaves as a Brønsted base, or can
donate a proton in water to yield H and SO42, thus behaving as a Brønsted acid.
Strategy: What are the characteristics of a Brønsted acid? Does it contain at least an H atom? With the
exception of ammonia, most Brønsted bases that you will encounter at this stage are anions.
Solution:
(a) PO43 in water can accept a proton to become HPO42, and is thus a Brønsted base.
(b)
ClO2 in water can accept a proton to become HClO2, and is thus a Brønsted base.
(c)
NH4 dissolved in water can donate a proton H, thus behaving as a Brønsted acid.
(d)
HCO3 can either accept a proton to become H2CO3, thus behaving as a Brønsted base. Or, HCO3
can donate a proton to yield H and CO32, thus behaving as a Brønsted acid.
Comment: The HCO3 species is said to be amphoteric because it possesses both acidic and basic
properties.
4.33
Recall that strong acids and strong bases are strong electrolytes. They are completely ionized in solution. An
ionic equation will show strong acids and strong bases in terms of their free ions. A net ionic equation shows
only the species that actually take part in the reaction.
(a)
Ionic: H(aq)  Br(aq)  NH3(aq) 
 NH4(aq)  Br(aq)
Net ionic: H(aq)  NH3(aq) 
 NH4(aq)
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
(b)
73
Ionic: 3Ba2(aq)  6OH(aq)  2H3PO4(aq) 
 Ba3(PO4)2(s)  6H2O(l)
Net ionic: 3Ba2(aq)  6OH(aq)  2H3PO4(aq) 
 Ba3(PO4)2(s)  6H2O(l)
(c)
Ionic: 2H(aq)  2ClO4(aq)  Mg2(aq)  2OH(aq) 
 Mg2(aq)  2ClO4(aq)  2H2O(l)
Net ionic: 2H(aq)  2OH(aq) 
 2H2O(l)
4.34
or
H(aq)  OH(aq) 
 H2O(l)
Strategy: Recall that strong acids and strong bases are strong electrolytes. They are completely ionized in
solution. An ionic equation will show strong acids and strong bases in terms of their free ions. Weak acids
and weak bases are weak electrolytes. They only ionize to a small extent in solution. Weak acids and weak
bases are shown as molecules in ionic and net ionic equations. A net ionic equation shows only the species
that actually take part in the reaction.
(a)
Solution: CH3COOH is a weak acid. It will be shown as a molecule in the ionic equation. KOH is a strong
base. It completely ionizes to K and OH ions. Since CH3COOH is an acid, it donates an H to the base,
OH, producing water. The other product is the salt, CH3COOK, which is soluble and remains in solution.
The balanced molecular equation is:
CH3COOH(aq)  KOH(aq) 
 CH3COOK(aq)  H2O(l)
The ionic and net ionic equations are:
Ionic: CH3COOH(aq)  K(aq)  OH(aq) 
 CH3COO(aq)  K(aq)  H2O(l)
Net ionic: CH3COOH(aq)  OH(aq) 
 CH3COO(aq)  H2O(l)
(b)
Solution: H2CO3 is a weak acid. It will be shown as a molecule in the ionic equation. NaOH is a strong
base. It completely ionizes to Na and OH ions. Since H2CO3 is an acid, it donates an H to the base, OH,
producing water. The other product is the salt, Na2CO3, which is soluble and remains in solution. The
balanced molecular equation is:
H2CO3(aq)  2NaOH(aq) 
 Na2CO3(aq)  2H2O(l)
The ionic and net ionic equations are:
Ionic: H2CO3(aq)  2Na(aq)  2OH(aq) 
 2Na(aq)  CO32(aq)  2H2O(l)
Net ionic: H2CO3(aq)  2OH(aq) 
 CO32(aq)  2H2O(l)
(c)
Solution: HNO3 is a strong acid. It completely ionizes to H and NO3 ions. Ba(OH)2 is a strong base. It
completely ionizes to Ba2 and OH ions. Since HNO3 is an acid, it donates an H to the base, OH,
producing water. The other product is the salt, Ba(NO 3)2, which is soluble and remains in solution. The
balanced molecular equation is:
2HNO3(aq)  Ba(OH)2(aq) 
 Ba(NO3)2(aq)  2H2O(l)
The ionic and net ionic equations are:
Ionic: 2H(aq)  2NO3(aq)  Ba2(aq)  2OH(aq) 
 Ba2(aq)  2NO3(aq)  2H2O(l)
Net ionic: 2H(aq)  2OH(aq) 
 2H2O(l)
or
H(aq)  OH(aq) 
 H2O(l)
74
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.43
Even though the problem doesn’t ask you to assign oxidation numbers, you need to be able to do so in order
to determine what is being oxidized or reduced.
(a)
(b)
(c)
(d)
4.44
(i) Half Reactions
(ii) Oxidizing Agent
(iii) Reducing Agent
Sr  Sr2  2e
O2  4e  2O2
O2
Sr
Li  Li  e
H2  2e  2H
H2
Li
Cs  Cs  e
Br2  2e  2Br
Br2
Cs
Mg  Mg2  2e
N2  6e  2N3
N2
Mg
Strategy: In order to break a redox reaction down into an oxidation half-reaction and a reduction halfreaction, you should first assign oxidation numbers to all the atoms in the reaction. In this way, you can
determine which element is oxidized (loses electrons) and which element is reduced (gains electrons).
Solution: In each part, the reducing agent is the reactant in the first half-reaction and the oxidizing agent is
the reactant in the second half-reaction. The coefficients in each half-reaction have been reduced to smallest
whole numbers.
(a)
The product is an ionic compound whose ions are Fe3 and O2.
Fe 
 Fe3  3e
O2  4e 
 2O2
O2 is the oxidizing agent; Fe is the reducing agent.
(b)
Na does not change in this reaction. It is a “spectator ion.”
2Br 
 Br2  2e
Cl2  2e 
 2Cl
Cl2 is the oxidizing agent; Br is the reducing agent.
(c)
Assume SiF4 is made up of Si4 and F.
Si 
 Si4  4e
F2  2e 
 2F
F2 is the oxidizing agent; Si is the reducing agent.
(d)
Assume HCl is made up of H and Cl.
H2 
 2H  2e
Cl2  2e 
 2Cl
Cl2 is the oxidizing agent; H2 is the reducing agent.
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.45
75
The oxidation number for hydrogen is 1 (rule 4), and for oxygen is 2 (rule 3). The oxidation number for
sulfur in S8 is zero (rule 1). Remember that in a neutral molecule, the sum of the oxidation numbers of all the
atoms must be zero, and in an ion the sum of oxidation numbers of all elements in the ion must equal the net
charge of the ion (rule 6).
H2S (2), S2 (2), HS (2) < S8 (0) < SO2 (4) < SO3 (6), H2SO4 (6)
The number in parentheses denotes the oxidation number of sulfur.
4.46
Strategy: In general, we follow the rules listed in Section 4.4 of the text for assigning oxidation numbers.
Remember that all alkali metals have an oxidation number of 1 in ionic compounds, and in most cases hydrogen
has an oxidation number of 1 and oxygen has an oxidation number of 2 in their compounds.
Solution: All the compounds listed are neutral compounds, so the oxidation numbers must sum to zero
(Rule 6, Section 4.4 of the text).
Let the oxidation number of P  x.
(a)
(b)
(c)
x  1  (3)(-2)  0, x  5
x  (3)(1)  (2)(-2)  0, x  1
x  (3)(1)  (3)(-2)  0, x  3
(d)
(e)
(f)
x  (3)(1)  (4)(-2)  0, x  5
2x  (4)(1)  (7)(-2)  0, 2x  10, x  5
3x  (5)(1)  (10)(-2)  0, 3x  15, x  5
The molecules in part (a), (e), and (f) can be made by strongly heating the compound in part (d). Are these
oxidation-reduction reactions?
Check: In each case, does the sum of the oxidation numbers of all the atoms equal the net charge on the
species, in this case zero?
4.47
See Section 4.4 of the text.
(a)
ClF: F 1 (rule 5), Cl 1 (rule 6)
(b)
IF7: F 1 (rule 5), I 7 (rules 5 and 6)
(c)
CH4: H 1 (rule 4), C 4 (rule 6)
(d)
C2H2: H 1 (rule 4), C 1 (rule 6)
(e)
C2H4: H 1 (rule 4), C 2 (rule 6),
(f)
K2CrO4: K 1 (rule 2), O 2 (rule 3), Cr 6 (rule 6)
(g)
K2Cr2O7: K 1 (rule 2), O 2 (rule 3), Cr 6 (rule 6)
(h)
KMnO4: K 1 (rule 2), O 2 (rule 3), Mn 7 (rule 6)
(i)
NaHCO3: Na 1 (rule 2), H 1 (rule 4), O 2 (rule 3), C 4 (rule 6)
(j)
Li2: Li 0 (rule 1)
(k)
(l)
KO2: K 1 (rule 2), O 1/2 (rule 6)
(m) PF6: F 1 (rule 5), P 5 (rule 6)
(n)
KAuCl4: K 1 (rule 2), Cl 1 (rule 5), Au 3 (rule 6)
NaIO3: Na 1 (rule 2), O 2 (rule 3), I 5 (rule 6)
4.48
All are free elements, so all have an oxidation number of zero.
4.49
(a) Cs2O, 1
(f) MoO42, 6
(k) SbF6, 5
(b) CaI2, 1
(g) PtCl42, 2
(c) Al2O3, 3
(h) PtCl62, 4
(d) H3AsO3, 3
(i) SnF2, 2
(e) TiO2, 4
(j) ClF3, 3
76
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.50
(a)
N: 3
(b)
O: 1/2
(c)
C: 1
(d)
C: 4
(e)
C: 3
(f)
O: 2
(g)
B: 3
(h)
W: 6
4.51
If nitric acid is a strong oxidizing agent and zinc is a strong reducing agent, then zinc metal will probably
reduce nitric acid when the two react; that is, N will gain electrons and the oxidation number of N must
decrease. Since the oxidation number of nitrogen in nitric acid is 5 (verify!), then the nitrogen-containing
product must have a smaller oxidation number for nitrogen. The only compound in the list that doesn’t have a
nitrogen oxidation number less than 5 is N2O5, (what is the oxidation number of N in N2O5?). This is never
a product of the reduction of nitric acid.
4.52
Strategy: Hydrogen displacement: Any metal above hydrogen in the activity series will displace it from
water or from an acid. Metals below hydrogen will not react with either water or an acid.
Solution: Only (b) Li and (d) Ca are above hydrogen in the activity series, so they are the only metals in
this problem that will react with water.
4.53
In order to work this problem, you need to assign the oxidation numbers to all the elements in the compounds.
In each case oxygen has an oxidation number of 2 (rule 3). These oxidation numbers should then be
compared to the range of possible oxidation numbers that each element can have. Molecular oxygen is a
powerful oxidizing agent. In SO3 alone, the oxidation number of the element bound to oxygen (S) is at its
maximum value (6); the sulfur cannot be oxidized further. The other elements bound to oxygen in this
problem have less than their maximum oxidation number and can undergo further oxidation.
4.54
(a)
Cu(s)  HCl(aq)  no reaction, since Cu(s) is less reactive than the hydrogen from acids.
(b)
I2(s)  NaBr(aq)  no reaction, since I2(s) is less reactive than Br2(l).
(c)
Mg(s)  CuSO4(aq)  MgSO4(aq)  Cu(s), since Mg(s) is more reactive than Cu(s).
Net ionic equation: Mg(s)  Cu2(aq)  Mg2(aq)  Cu(s)
(d)
Cl2(g)  2KBr(aq)  Br2(l)  2KCl(aq), since Cl2(g) is more reactive than Br2(l)
Net ionic equation: Cl2(g)  2Br(aq)  2Cl(aq)  Br2(l)
4.55
(a)
(c)
Disproportionation reaction
Decomposition reaction
4.56
(a)
(b)
(c)
(d)
Combination reaction
Decomposition reaction
Displacement reaction
Disproportionation reaction
4.59
First, calculate the moles of KI needed to prepare the solution.
mol KI 
(b)
(d)
Displacement reaction
Combination reaction
2.80 mol KI
 (5.00  102 mL soln)  1.40 mol KI
1000 mL soln
Converting to grams of KI:
1.40 mol KI 
166.0 g KI
 232 g KI
1 mol KI
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.60
77
Strategy: How many moles of NaNO3 does 250 mL of a 0.707 M solution contain? How would you
convert moles to grams?
Solution: From the molarity (0.707 M), we can calculate the moles of NaNO3 needed to prepare 250 mL of
solution.
0.707 mol NaNO3
Moles NaNO3 
 250 mL soln  0.177 mol
1000 mL soln
Next, we use the molar mass of NaNO3 as a conversion factor to convert from moles to grams.
M (NaNO3)  85.00 g/mol.
0.177 mol NaNO3 
85.00 g NaNO3
 15.0 g NaNO3
1 mol NaNO3
To make the solution, dissolve 15.0 g of NaNO3 in enough water to make 250 mL of solution.
Check: As a ball-park estimate, the mass should be given by [molarity (mol/L)  volume (L)  moles 
molar mass (g/mol)  grams]. Let's round the molarity to 1 M and the molar mass to 80 g, because we are
simply making an estimate. This gives: [1 mol/L  (1/4)L  80 g  20 g]. This is close to our answer of
15.0 g.
4.61
mol  M  L
60.0 mL  0.0600 L
mol MgCl 2 
4.62
0.100 mol MgCl2
 0.0600 L soln  6.00  103 mol MgCl 2
1 L soln
Since the problem asks for grams of solute (KOH), you should be thinking that you can calculate moles of
solute from the molarity and volume of solution. Then, you can convert moles of solute to grams of solute.
? moles KOH solute 
5.50 moles solute
 35.0 mL solution  0.193 mol KOH
1000 mL solution
The molar mass of KOH is 56.11 g/mol. Use this conversion factor to calculate grams of KOH.
? grams KOH  0.193 mol KOH 
4.63
56.11 g KOH
 10.8 g KOH
1 mol KOH
Molar mass of C2H5OH  46.07 g/mol; molar mass of C12H22O11  342.3 g/mol; molar mass of
NaCl  58.44 g/mol.
(a)
? mol C2 H5OH  29.0 g C2 H5OH 
Molarity 
(b)
1 mol C2 H5OH
 0.629 mol C2 H5OH
46.07 g C2 H5OH
0.629 mol C2 H5OH
mol solute

 1.15 M
L of soln
0.545 L soln
? mol C12 H 22O11  15.4 g C12 H 22O11 
1 mol C12 H 22O11
 0.0450 mol C12H 22O11
342.3 g C12 H 22O11
78
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
Molarity 
(c)
? mol NaCl  9.00 g NaCl 
Molarity 
4.64
(a)
1 mol CH3OH
 0.205 mol CH3OH
32.04 g CH3OH
0.205 mol CH3OH
 1.37 M
0.150 L
1 mol CaCl2
 0.0937 mol CaCl2
111.0 g CaCl2
0.0937 mol CaCl2
 0.426 M
0.220 L
? mol C10 H8  7.82 g C10 H8 
M 
4.65
mol solute
0.154 mol NaCl

 1.78 M
L of soln
86.4  103 L soln
? mol CaCl2  10.4 g CaCl2 
M 
(c)
1 mol NaCl
 0.154 mol NaCl
58.44 g NaCl
? mol CH3OH  6.57 g CH3OH 
M 
(b)
0.0450 mol C12 H 22 O11
mol solute

 0.608 M
L of soln
74.0  103 L soln
1 mol C10 H8
 0.0610 mol C10 H8
128.2 g C10 H8
0.0610 mol C10 H8
 0.716 M
0.0852 L
First, calculate the moles of each solute. Then, you can calculate the volume (in L) from the molarity and the
number of moles of solute.
(a)
? mol NaCl  2.14 g NaCl 
L soln 
(b)
mol solute
0.0366 mol NaCl

 0.136 L  136 mL soln
Molarity
0.270 mol/L
? mol C2 H5OH  4.30 g C2 H5OH 
L soln 
(c)
1 mol NaCl
 0.0366 mol NaCl
58.44 g NaCl
1 mol C2 H5OH
 0.0933 mol C2 H5OH
46.07 g C2 H5OH
0.0933 mol C2 H5OH
mol solute

 0.0622 L  62.2 mL soln
Molarity
1.50 mol/L
? mol CH3COOH  0.85 g CH3COOH 
L soln 
1 mol CH3COOH
 0.014 mol CH3COOH
60.05 g CH3COOH
0.014 mol CH3COOH
mol solute

 0.047 L  47 mL soln
Molarity
0.30 mol/L
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.66
A 250 mL sample of 0.100 M solution contains 0.0250 mol of solute (mol  M  L). The computation in each
case is the same:
259.8 g CsI
 6.50 g CsI
1 mol CsI
(a)
0.0250 mol CsI 
(b)
0.0250 mol H 2SO4 
(c)
0.0250 mol Na 2CO3 
(d)
0.0250 mol K 2 Cr2 O7 
(e) 0.0250 mol KMnO4 
4.69
79
98.09 g H 2SO4
 2.45 g H 2SO4
1 mol H 2SO4
106.0 g Na 2CO3
 2.65 g Na2CO3
1 mol Na 2 CO3
294.2 g K 2 Cr2 O7
 7.36 g K 2Cr2O7
1 mol K 2 Cr2 O7
158.0 g KMnO4
 3.95 g KMnO4
1 mol KMnO4
MinitialVinitial  MfinalVfinal
You can solve the equation algebraically for Vinitial. Then substitute in the given quantities to solve for the
volume of 2.00 M HCl needed to prepare 1.00 L of a 0.646 M HCl solution.
Vinitial 
M final  Vfinal
0.646 M  1.00 L

 0.323 L  323 mL
M initial
2.00 M
To prepare the 0.646 M solution, you would dilute 323 mL of the 2.00 M HCl solution to a final volume of
1.00 L.
4.70
Strategy: Because the volume of the final solution is greater than the original solution, this is a dilution
process. Keep in mind that in a dilution, the concentration of the solution decreases, but the number of moles
of the solute remains the same.
Solution: We prepare for the calculation by tabulating our data.
Mi  0.866 M
Mf  ?
Vi  25.0 mL
Vf  500 mL
We substitute the data into Equation (4.2) of the text.
MiVi  MfVf
(0.866 M)(25.0 mL)  Mf(500 mL)
Mf 
4.71
(0.866 M )(25.0 mL)
 0.0433 M
500 mL
MinitialVinitial  MfinalVfinal
You can solve the equation algebraically for Vinitial. Then substitute in the given quantities to solve the for the
volume of 4.00 M HNO3 needed to prepare 60.0 mL of a 0.200 M HNO3 solution.
80
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
Vinitial 
M final  Vfinal
0.200 M  60.00 mL

 3.00 mL
M initial
4.00 M
To prepare the 0.200 M solution, you would dilute 3.00 mL of the 4.00 M HNO3 solution to a final volume of
60.0 mL.
4.72
You need to calculate the final volume of the dilute solution. Then, you can subtract 505 mL from this
volume to calculate the amount of water that should be added.
Vfinal 
 0.125 M  505 mL 
M initialVinitial

 631 mL
M final
 0.100 M 
(631  505) mL  126 mL of water
4.73
Moles of KMnO4 in the first solution:
1.66 mol
 35.2 mL  0.0584 mol KMnO4
1000 mL soln
Moles of KMnO4 in the second solution:
0.892 mol
 16.7 mL  0.0149 mol KMnO4
1000 mL soln
The total volume is 35.2 mL  16.7 mL  51.9 mL. The concentration of the final solution is:
M 
4.74
 0.0584  0.0149  mol
51.9  103 L
 1.41 M
Moles of calcium nitrate in the first solution:
0.568 mol
 46.2 mL soln  0.0262 mol Ca(NO3 ) 2
1000 mL soln
Moles of calcium nitrate in the second solution:
1.396 mol
 80.5 mL soln  0.112 mol Ca(NO3 ) 2
1000 mL soln
The volume of the combined solutions  46.2 mL  80.5 mL  126.7 mL. The concentration of the final
solution is:
(0.0262  0.112) mol
M 
 1.09 M
0.1267 L
4.77
The balanced equation is:
CaCl2(aq)  2AgNO3(aq) 
 Ca(NO3)2(aq)  2AgCl(s)
We need to determine the limiting reagent. Ag and Cl combine in a 1:1 mole ratio to produce AgCl. Let’s
calculate the amount of Ag and Cl in solution.
mol Ag  
0.100 mol Ag 
 15.0 mL soln  1.50  103 mol Ag 
1000 mL soln
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
mol Cl 
81
0.150 mol CaCl2
2 mol Cl

 30.0 mL soln  9.00  103 mol Cl
1000 mL soln
1 mol CaCl2
Since Ag and Cl combine in a 1:1 mole ratio, AgNO3 is the limiting reagent. Only 1.50  103 mole of
AgCl can form. Converting to grams of AgCl:
1.50  103 mol AgCl 
4.78
143.4 g AgCl
 0.215 g AgCl
1 mol AgCl
Strategy: We want to calculate the mass % of Ba in the original compound. Let's start with the definition of
mass %.
need to find
want to calculate
mass % Ba 
mass Ba
 100%
mass of sample
given
The mass of the sample is given in the problem (0.6760 g). Therefore we need to find the mass of Ba in the
original sample. We assume the precipitation is quantitative, that is, that all of the barium in the sample has
been precipitated as barium sulfate. From the mass of BaSO4 produced, we can calculate the mass of Ba.
There is 1 mole of Ba in 1 mole of BaSO4.
Solution: First, we calculate the mass of Ba in 0.4105 g of the BaSO4 precipitate. The molar mass of
BaSO4 is 233.4 g/mol.
1 mol BaSO4
1 mol Ba
137.3 g Ba


233.4 g BaSO4 1 mol BaSO4
1 mol Ba
? mass of Ba  0.4105 g BaSO4 
 0.2415 g Ba
Next, we calculate the mass percent of Ba in the unknown compound.
%Ba by mass 
4.79
0.2415 g
 100%  35.72%
0.6760 g
The net ionic equation is: Ag(aq)  Cl(aq) 
 AgCl(s)
One mole of Cl is required per mole of Ag. First, find the number of moles of Ag.
mol Ag  
0.0113 mol Ag 
 (2.50  102 mL soln)  2.83  103 mol Ag 
1000 mL soln
Now, calculate the mass of NaCl using the mole ratio from the balanced equation.
(2.83  103 mol Ag  ) 
1 mol Cl
1 mol Ag


1 mol NaCl
1 mol Cl


58.44 g NaCl
 0.165 g NaCl
1 mol NaCl
82
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.80
The net ionic equation is:
Cu2(aq)  S2(aq) 
 CuS(s)
The answer sought is the molar concentration of Cu2, that is, moles of Cu2 ions per liter of solution. The
factor-label method is used to convert, in order:
g of CuS  moles CuS  moles Cu2  moles Cu2 per liter soln
[Cu2+ ]  0.0177 g CuS 
4.85
1 mol CuS
1 mol Cu 2
1


 2.31  104 M
95.62 g CuS 1 mol CuS 0.800 L
The reaction between KHP (KHC8H4O4) and KOH is:
KHC8H4O4(aq)  KOH(aq)  H2O(l)  K2C8H4O4(aq)
We know the volume of the KOH solution, and we want to calculate the molarity of the KOH solution.
want to calculate
M of KOH 
need to find
mol KOH
L of KOH soln
given
If we can determine the moles of KOH in the solution, we can then calculate the molarity of the solution.
From the mass of KHP and its molar mass, we can calculate moles of KHP. Then, using the mole ratio from
the balanced equation, we can calculate moles of KOH.
? mol KOH  0.4218 g KHP 
1 mol KHP
1 mol KOH

 2.066  103 mol KOH
204.2 g KHP 1 mol KHP
From the moles and volume of KOH, we calculate the molarity of the KOH solution.
M of KOH 
4.86
mol KOH
2.066  103 mol KOH

 0.1106 M
L of KOH soln
18.68  103 L soln
The reaction between HCl and NaOH is:
HCl(aq)  NaOH(aq)  H2O(l)  NaCl(aq)
We know the volume of the NaOH solution, and we want to calculate the molarity of the NaOH solution.
want to calculate
M of NaOH 
need to find
mol NaOH
L of NaOH soln
given
If we can determine the moles of NaOH in the solution, we can then calculate the molarity of the solution.
From the volume and molarity of HCl, we can calculate moles of HCl. Then, using the mole ratio from the
balanced equation, we can calculate moles of NaOH.
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
? mol NaOH  17.4 mL HCl 
83
0.312 mol HCl 1 mol NaOH

 5.43  103 mol NaOH
1000 mL soln
1 mol HCl
From the moles and volume of NaOH, we calculate the molarity of the NaOH solution.
M of NaOH 
4.87
(a)
mol NaOH
5.43  103 mol NaOH

 0.217 M
L of NaOH soln
25.0  103 L soln
In order to have the correct mole ratio to solve the problem, you must start with a balanced chemical
equation.
HCl(aq)  NaOH(aq) 
 NaCl(aq)  H2O(l)
From the molarity and volume of the HCl solution, you can calculate moles of HCl. Then, using the
mole ratio from the balanced equation above, you can calculate moles of NaOH.
? mol NaOH  25.00 mL 
2.430 mol HCl 1 mol NaOH

 6.075  102 mol NaOH
1000 mL soln
1 mol HCl
Solving for the volume of NaOH:
liters of solution 
volume of NaOH 
(b)
moles of solute
M
6.075  102 mol NaOH
 4.278  102 L  42.78 mL
1.420 mol/L
This problem is similar to part (a). The difference is that the mole ratio between base and acid is 2:1.
H2SO4(aq)  2NaOH(aq) 
 Na2SO4(aq)  H2O(l)
? mol NaOH  25.00 mL 
volume of NaOH 
(c)
4.500 mol H2SO4 2 mol NaOH

 0.2250 mol NaOH
1000 mL soln
1 mol H 2SO4
0.2250 mol NaOH
 0.1585 L  158.5 mL
1.420 mol/L
This problem is similar to parts (a) and (b). The difference is that the mole ratio between base and acid
is 3:1.
H3PO4(aq)  3NaOH(aq) 
 Na3PO4(aq)  3H2O(l)
? mol NaOH  25.00 mL 
volume of NaOH 
1.500 mol H3PO4 3 mol NaOH

 0.1125 mol NaOH
1000 mL soln
1 mol H3PO4
0.1125 mol NaOH
 0.07923 L  79.23 mL
1.420 mol/L
84
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.88
Strategy: We know the molarity of the HCl solution, and we want to calculate the volume of the HCl
solution.
need to find
given
M of HCl 
mol HCl
L of HCl soln
want to calculate
If we can determine the moles of HCl, we can then use the definition of molarity to calculate the volume of
HCl needed. From the volume and molarity of NaOH or Ba(OH) 2, we can calculate moles of NaOH or
Ba(OH)2. Then, using the mole ratio from the balanced equation, we can calculate moles of HCl.
Solution:
(a)
In order to have the correct mole ratio to solve the problem, you must start with a balanced chemical
equation.
HCl(aq)  NaOH(aq) 
 NaCl(aq)  H2O(l)
? mol HCl  10.0 mL 
0.300 mol NaOH
1 mol HCl

 3.00  103 mol HCl
1000 mL of solution 1 mol NaOH
From the molarity and moles of HCl, we calculate volume of HCl required to neutralize the NaOH.
(b)
liters of solution 
moles of solute
M
volume of HCl 
3.00  103 mol HCl
 6.00  103 L  6.00 mL
0.500 mol/L
This problem is similar to part (a). The difference is that the mole ratio between acid and base is 2:1.
2HCl(aq)  Ba(OH)2(aq) 
 BaCl2(aq) 2H2O(l)
? mol HCl  10.0 mL 
volume of HCl 
4.91
0.200 mol Ba(OH)2
2 mol HCl

 4.00  103 mol HCl
1000 mL of solution 1 mol Ba(OH)2
4.00  103 mol HCl
 8.00  103 L = 8.00 mL
0.500 mol/L
The balanced equation is given in the problem. The mole ratio between Fe2 and Cr2O72 is 6:1.
First, calculate the moles of Fe2 that react with Cr2O72.
26.00 mL soln 
0.0250 mol Cr2O72
6 mol Fe2

 3.90  103 mol Fe2
1000 mL soln
1 mol Cr2O72
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
85
The molar concentration of Fe2 is:
M 
4.92
3.90  103 mol Fe2
25.0  103 L soln
 0.156 M
Strategy: We want to calculate the grams of SO2 in the sample of air. From the molarity and volume of
KMnO4, we can calculate moles of KMnO4. Then, using the mole ratio from the balanced equation, we can
calculate moles of SO2. How do we convert from moles of SO2 to grams of SO2?
Solution: The balanced equation is given in the problem.
5SO2  2MnO4  2H2O 
 5SO42  2Mn2  4H
The moles of KMnO4 required for the titration are:
0.00800 mol KMnO4
 7.37 mL  5.90  105 mol KMnO4
1000 mL soln
We use the mole ratio from the balanced equation and the molar mass of SO 2 as conversion factors to convert
to grams of SO2.
(5.90  105 mol KMnO4 ) 
4.93
5 mol SO2
64.07 g SO2

 9.45  103 g SO2
2 mol KMnO4
1 mol SO2
The balanced equation is given in problem 4.91. The mole ratio between Fe 2 and Cr2O72 is 6:1.
First, calculate the moles of Cr2O72 that reacted.
23.30 mL soln 
0.0194 mol Cr2 O72
 4.52  104 mol Cr2 O72
1000 mL soln
Use the mole ratio from the balanced equation to calculate the mass of iron that reacted.
(4.52  104 mol Cr2 O72 ) 
6 mol Fe2
1 mol Cr2 O72

55.85 g Fe2
1 mol Fe 2
 0.151 g Fe 2
The percent by mass of iron in the ore is:
0.151 g
 100%  54.1%
0.2792 g
4.94
The balanced equation is given in the problem.
2MnO4  5H2O2  6H 
 5O2  2Mn2  8H2O
First, calculate the moles of potassium permanganate in 36.44 mL of solution.
0.01652 mol KMnO4
 36.44 mL  6.020  104 mol KMnO4
1000 mL soln
86
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
Next, calculate the moles of hydrogen peroxide using the mole ratio from the balanced equation.
(6.020  104 mol KMnO4 ) 
5 mol H 2O2
 1.505  103 mol H 2O 2
2 mol KMnO4
Finally, calculate the molarity of the H2O2 solution. The volume of the solution is 0.02500 L.
Molarity of H 2O2 
4.95
1.505  103 mol H2 O2
 0.06020 M
0.02500 L
The balanced equation shows that 2 moles of electrons are lost for each mole of SO 32 that reacts. The
electrons are gained by IO3. We need to find the moles of electrons gained for each mole of IO 3 that reacts.
Then, we can calculate the final oxidation state of iodine.
The number of moles of electrons lost by SO32 is:
32.5 mL 
0.500 mol SO32
2 mol e

 0.0325 mol e lost
2
1000 mL soln
1 mol SO3
The number of moles of iodate, IO3, that react is:
1.390 g KIO3 
1 mol KIO3
1 mol IO3

 6.495  103 mol IO3
214.0 g KIO3 1 mol KIO3
6.495  103 mole of IO3 gain 0.0325 mole of electrons. The number of moles of electrons gained per mole
of IO3 is:
0.0325 mol e 
6.495  103 mol IO3
 5.00 mol e  /mol IO3
The oxidation number of iodine in IO3 is 5. Since 5 moles of electrons are gained per mole of IO 3, the
final oxidation state of iodine is 5  5  0. The iodine containing product of the reaction is most likely
elemental iodine, I2.
4.96
First, calculate the moles of KMnO4 in 24.0 mL of solution.
0.0100 mol KMnO4
 24.0 mL  2.40  104 mol KMnO4
1000 mL soln
Next, calculate the mass of oxalic acid needed to react with 2.40  104 mol KMnO4. Use the mole ratio
from the balanced equation.
(2.40  104 mol KMnO4 ) 
5 mol H 2 C2 O4 90.04 g H 2 C2 O4

 0.0540 g H 2C2O 4
2 mol KMnO4
1 mol H 2C2O4
The original sample had a mass of 1.00 g. The mass percent of H2C2O4 in the sample is:
mass % 
0.0540 g
 100%  5.40% H 2C2O4
1.00 g
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.97
87
The first titration oxidizes Fe2 to Fe3. This titration gives the amount of Fe2 in solution. Zn metal is
added to reduce all Fe3 back to Fe2. The second titration oxidizes all the Fe2 back to Fe3. We can find
the amount of Fe3 in the original solution by difference.
Titration #1: The mole ratio between Fe2 and MnO4 is 5:1.
23.0 mL soln 
[Fe2+ ] 
0.0200 mol MnO4
5 mol Fe2

 2.30  103 mol Fe2

1000 mL soln
1 mol MnO4
mol solute
2.30  103 mol Fe 2

 0.0920 M
L of soln
25.0  103 L soln
Titration #2: The mole ratio between Fe2 and MnO4 is 5:1.
40.0 mL soln 
0.0200 mol MnO4
5 mol Fe2

 4.00  103 mol Fe2

1000 mL soln
1 mol MnO4
In this second titration, there are more moles of Fe2 in solution. This is due to Fe3 in the original solution
being reduced by Zn to Fe2. The number of moles of Fe3 in solution is:
(4.00  103 mol)  (2.30  103 mol)  1.70  103 mol Fe3
[Fe3+ ] 
4.98
mol solute
1.70  103 mol Fe3

 0.0680 M
L of soln
25.0  103 L soln
The balanced equation is:
2MnO4  16H  5C2O42 
 2Mn2  10CO2  8H2O
mol MnO4 
9.56  104 mol MnO4
 24.2 mL  2.31  105 mol MnO4
1000 mL of soln
Using the mole ratio from the balanced equation, we can calculate the mass of Ca 2 in the 10.0 mL sample of
blood.
(2.31  105 mol MnO4 ) 
5 mol C2O24
2 mol MnO4

1 mol Ca 2
1 mol C2O42

40.08 g Ca 2
1 mol Ca 2
 2.31  103 g Ca 2
Converting to mg/mL:
2.31  103 g Ca 2
1 mg

 0.231 mg Ca2+ /mL of blood
10.0 mL of blood
0.001 g
4.99
In redox reactions the oxidation numbers of elements change. To test whether an equation represents a redox
process, assign the oxidation numbers to each of the elements in the reactants and products. If oxidation
numbers change, it is a redox reaction.
(a)
On the left the oxidation number of chlorine in Cl2 is zero (rule 1). On the right it is 1 in Cl (rule 2)
and 1 in OCl (rules 3 and 5). Since chlorine is both oxidized and reduced, this is a disproportionation
redox reaction.
88
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
(b)
The oxidation numbers of calcium and carbon do not change. This is not a redox reaction; it is a
precipitation reaction.
(c)
The oxidation numbers of nitrogen and hydrogen do not change. This is not a redox reaction; it is an
acid-base reaction.
(d)
The oxidation numbers of carbon, chlorine, chromium, and oxygen do not change. This is not a redox
reaction; it doesn’t fit easily into any category, but could be considered as a type of combination
reaction.
(e)
The oxidation number of calcium changes from 0 to 2, and the oxidation number of fluorine changes
from 0 to 1. This is a combination redox reaction.
The remaining parts (f) through (j) can be worked the same way.
(f)
Redox
(g)
Precipitation
(h)
Redox
(i)
Redox
(j)
Redox
4.100
First, the gases could be tested to see if they supported combustion. O2 would support combustion, CO2
would not. Second, if CO2 is bubbled through a solution of calcium hydroxide [Ca(OH) 2], a white precipitate
of CaCO3 forms. No reaction occurs when O2 is bubbled through a calcium hydroxide solution.
4.101
Choice (d), 0.20 M Mg(NO3)2, should be the best conductor of electricity; the total ion concentration in this
solution is 0.60 M. The total ion concentrations for solutions (a) and (c) are 0.40 M and 0.50 M, respectively.
We can rule out choice (b), because acetic acid is a weak electrolyte.
4.102
Starting with a balanced chemical equation:
Mg(s)  2HCl(aq) 
 MgCl2(aq)  H2(g)
From the mass of Mg, you can calculate moles of Mg. Then, using the mole ratio from the balanced equation
above, you can calculate moles of HCl reacted.
4.47 g Mg 
1 mol Mg
2 mol HCl

 0.368 mol HCl reacted
24.31 g Mg 1 mol Mg
Next we can calculate the number of moles of HCl in the original solution.
2.00 mol HCl
 (5.00  102 mL)  1.00 mol HCl
1000 mL soln
Moles HCl remaining  1.00 mol  0.368 mol  0.632 mol HCl
conc. of HCl after reaction 
4.103
mol HCl
0.632 mol HCl

 1.26 mol/L  1.26 M
L soln
0.500 L
The balanced equation for the displacement reaction is:
Zn(s)  CuSO4(aq) 
 ZnSO4(aq)  Cu(s)
The moles of CuSO4 that react with 7.89 g of zinc are:
7.89 g Zn 
1 mol CuSO4
1 mol Zn

 0.121 mol CuSO4
65.39 g Zn
1 mol Zn
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
89
The volume of the 0.156 M CuSO4 solution needed to react with 7.89 g Zn is:
L of soln 
0.121 mol CuSO4
mole solute

 0.776 L  776 mL
M
0.156 mol/L
Would you expect Zn to displace Cu2 from solution, as shown in the equation?
4.104
The balanced equation is:
2HCl(aq)  Na2CO3(s) 
 CO2(g)  H2O(l)  2NaCl(aq)
The mole ratio from the balanced equation is 2 moles HCl : 1 mole Na 2CO3. The moles of HCl needed to
react with 0.256 g of Na2CO3 are:
0.256 g Na 2CO3 
Molarity HCl 
4.105
1 mol Na 2CO3
2 mol HCl

 4.83  103 mol HCl
106.0 g Na 2CO3 1 mol Na 2CO3
moles HCl
4.83  103 mol HCl

 0.171 mol/L  0.171 M
L soln
0.0283 L soln
The neutralization reaction is: HA(aq)  NaOH(aq) 
 NaA(aq)  H2O(l)
The mole ratio between the acid and NaOH is 1:1. The moles of HA that react with NaOH are:
20.27 mL soln 
0.1578 mol NaOH
1 mol HA

 3.199  103 mol HA
1000 mL soln
1 mol NaOH
3.664 g of the acid reacted with the base. The molar mass of the acid is:
Molar mass 
4.106
3.664 g HA
3.199  103 mol HA
 1145 g/mol
Starting with a balanced chemical equation:
CH3COOH(aq)  NaOH(aq) 
 CH3COONa(aq)  H2O(l)
From the molarity and volume of the NaOH solution, you can calculate moles of NaOH. Then, using the
mole ratio from the balanced equation above, you can calculate moles of CH 3COOH.
5.75 mL solution 
1 mol CH3COOH
1.00 mol NaOH

 5.75  103 mol CH3COOH
1000 mL of solution
1 mol NaOH
Molarity CH 3COOH 
4.107
5.75  103 mol CH3COOH
 0.115 M
0.0500 L
Let’s call the original solution, soln 1; the first dilution, soln 2; and the second dilution, soln 3. Start with the
concentration of soln 3, 0.00383 M. From the concentration and volume of soln 3, we can find the
concentration of soln 2. Then, from the concentration and volume of soln 2, we can find the concentration of
soln 1, the original solution.
M2V2  M3V3
90
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
M2 
M 3V3
(0.00383 M )(1.000  103 mL)

 0.153 M
V2
25.00 mL
M1 
M 2V2
(0.153 M )(125.0 mL)

 1.28 M
V1
15.00 mL
M1V1  M2V2
4.108
The balanced equation is:
Zn(s)  2AgNO3(aq) 
 Zn(NO3)2(aq)  2Ag(s)
Let x  mass of Ag produced. We can find the mass of Zn reacted in terms of the amount of Ag produced.
x g Ag 
1 mol Ag
1 mol Zn 65.39 g Zn


 0.303x g Zn reacted
107.9 g Ag 2 mol Ag
1 mol Zn
The mass of Zn remaining will be:
2.50 g  amount of Zn reacted  2.50 g Zn  0.303x g Zn
The final mass of the strip, 3.37 g, equals the mass of Ag produced  the mass of Zn remaining.
3.37 g  x g Ag  (2.50 g Zn  0.303 x g Zn)
x  1.25 g  mass of Ag produced
mass of Zn remaining  3.37 g  1.25 g  2.12 g Zn
or
mass of Zn remaining  2.50 g Zn  0.303x g Zn  2.50 g  (0.303)(1.25 g)  2.12 g Zn
4.109
The balanced equation is:
Ba(OH)2(aq)  Na2SO4(aq) 
 BaSO4(s)  2NaOH(aq)
moles Ba(OH)2: (2.27 L)(0.0820 mol/L)  0.186 mol Ba(OH)2
moles Na2SO4: (3.06 L)(0.0664 mol/L)  0.203 mol Na2SO4
Since the mole ratio between Ba(OH)2 and Na2SO4 is 1:1, Ba(OH)2 is the limiting reagent. The mass of
BaSO4 formed is:
1 mol BaSO4
233.4 g BaSO4
0.186 mol Ba(OH)2 

 43.4 g BaSO4
1 mol Ba(OH)2
mol BaSO4
4.110
The balanced equation is:
HNO3(aq)  NaOH(aq) 
 NaNO3(aq)  H2O(l)
mol HNO3 
0.211 mol HNO3
 10.7 mL soln  2.26  103 mol HNO3
1000 mL soln
mol NaOH 
0.258 mol NaOH
 16.3 mL soln  4.21  103 mol NaOH
1000 mL soln
Since the mole ratio from the balanced equation is 1 mole NaOH : 1 mole HNO 3, then 2.26  103 mol HNO3
will react with 2.26  103 mol NaOH.
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
91
mol NaOH remaining  (4.21  103 mol)  (2.26  103 mol)  1.95  103 mol NaOH
10.7 mL  16.3 mL  27.0 mL  0.027 L
molarity NaOH 
4.111
(a)
1.95  103 mol NaOH
 0.0722 M
0.027 L
Magnesium hydroxide is insoluble in water. It can be prepared by mixing a solution containing Mg 2
ions such as MgCl2(aq) or Mg(NO3)2(aq) with a solution containing hydroxide ions such as NaOH(aq).
Mg(OH)2 will precipitate, which can then be collected by filtration. The net ionic reaction is:
Mg2(aq)  2OH(aq)  Mg(OH)2(s)
(b)
The balanced equation is:
2HCl  Mg(OH)2 
 MgCl2  2H2O
The moles of Mg(OH)2 in 10 mL of milk of magnesia are:
10 mL soln 
0.080 g Mg(OH)2
1 mol Mg(OH) 2

 0.014 mol Mg(OH)2
1 mL soln
58.33 g Mg(OH)2
Moles of HCl reacted  0.014 mol Mg(OH) 2 
Volume of HCl 
4.112
2 mol HCl
 0.028 mol HCl
1 mol Mg(OH)2
mol solute
0.028 mol HCl

 0.80 L
M
0.035 mol/L
The balanced equations for the two reactions are:
X(s)  H2SO4(aq) 
 XSO4(aq)  H2(g)
H2SO4(aq)  2NaOH(aq) 
 Na2SO4(aq)  2H2O(l)
First, let’s find the number of moles of excess acid from the reaction with NaOH.
0.0334 L 
0.500 mol NaOH 1 mol H 2SO4

 8.35  103 mol H2SO4
1 L soln
2 mol NaOH
The original number of moles of acid was:
0.100 L 
0.500 mol H 2SO4
 0.0500 mol H 2SO4
1 L soln
The amount of sulfuric acid that reacted with the metal, X, is
(0.0500 mol H2SO4)  (8.35  103 mol H2SO4)  0.0417 mol H2SO4.
Since the mole ratio from the balanced equation is 1 mole X : 1 mole H2SO4, then the amount of X that
reacted is 0.0417 mol X.
1.00 g X
molar mass X 
 24.0 g/mol
0.0417 mol X
The element is magnesium.
92
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.113
Add a known quantity of compound in a given quantity of water. Filter and recover the undissolved
compound, then dry and weigh it. The difference in mass between the original quantity and the recovered
quantity is the amount that dissolved in the water.
4.114
First, calculate the number of moles of glucose present.
0.513 mol glucose
 60.0 mL  0.0308 mol glucose
1000 mL soln
2.33 mol glucose
 120.0 mL  0.280 mol glucose
1000 mL soln
Add the moles of glucose, then divide by the total volume of the combined solutions to calculate the molarity.
60.0 mL  120.0 mL  180.0 mL  0.180 L
Molarity of final solution 
(0.0308  0.280) mol glucose
 1.73 mol/L  1.73 M
0.180 L
4.115
First, you would accurately measure the electrical conductance of pure water. The conductance of a solution
of the slightly soluble ionic compound X should be greater than that of pure water. The increased
conductance would indicate that some of the compound X had dissolved.
4.116
Iron(II) compounds can be oxidized to iron(III) compounds. The sample could be tested with a small amount
of a strongly colored oxidizing agent like a KMnO4 solution, which is a deep purple color. A loss of color
would imply the presence of an oxidizable substance like an iron(II) salt.
4.117
The three chemical tests might include:
(1)
(2)
(3)
4.118
Electrolysis to ascertain if hydrogen and oxygen were produced,
The reaction with an alkali metal to see if a base and hydrogen gas were produced, and
The dissolution of a metal oxide to see if a base was produced (or a nonmetal oxide to see if an acid was
produced).
Since both of the original solutions were strong electrolytes, you would expect a mixture of the two solutions
to also be a strong electrolyte. However, since the light dims, the mixture must contain fewer ions than the
original solution. Indeed, H from the sulfuric acid reacts with the OH from the barium hydroxide to form
water. The barium cations react with the sulfate anions to form insoluble barium sulfate.
2H(aq)  SO42(aq)  Ba2(aq)  2OH(aq) 
 2H2O(l)  BaSO4(s)
Thus, the reaction depletes the solution of ions and the conductivity decreases.
4.119
(a)
Check with litmus paper, react with carbonate or bicarbonate to see if CO2 gas is produced, react with a
base and check with an indicator.
(b)
Titrate a known quantity of acid with a standard NaOH solution. Since it is a monoprotic acid, the
moles of NaOH reacted equals the moles of the acid. Dividing the mass of acid by the number of moles
gives the molar mass of the acid.
(c)
Visually compare the conductivity of the acid with a standard NaCl solution of the same molar
concentration. A strong acid will have a similar conductivity to the NaCl solution. The conductivity of
a weak acid will be considerably less than the NaCl solution.
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
93
4.120
You could test the conductivity of the solutions. Sugar is a nonelectrolyte and an aqueous sugar solution will
not conduct electricity; whereas, NaCl is a strong electrolyte when dissolved in water. Silver nitrate could be
added to the solutions to see if silver chloride precipitated. In this particular case, the solutions could also be
tasted.
4.121
(a)
Pb(NO3)2(aq)  Na2SO4(aq) 
 PbSO4(s)  2NaNO3(aq)
Pb2(aq)  SO42(aq) 
 PbSO4(s)
(b)
First, calculate the moles of Pb2 in the polluted water.
0.00450 g Na 2SO4 
1 mol Pb(NO3 )2
1 mol Na 2SO4
1 mol Pb2


 3.17  105 mol Pb 2
142.1 g Na 2SO4
1 mol Na 2SO4
1 mol Pb(NO3 ) 2
The volume of the polluted water sample is 500 mL (0.500 L). The molar concentration of Pb 2 is:
[Pb2+ ] 
mol Pb2
3.17  105 mol Pb2

 6.34  105 M
L of soln
0.500 L soln
4.122
In a redox reaction, the oxidizing agent gains one or more electrons. In doing so, the oxidation number of the
element gaining the electrons must become more negative. In the case of chlorine, the 1 oxidation number is
already the most negative state possible. The chloride ion cannot accept any more electrons; therefore,
hydrochloric acid is not an oxidizing agent.
4.123
(a)
An acid and a base react to form water and a salt. Potassium iodide is a salt; therefore, the acid and base
are chosen to produce this salt.
KOH(aq)  HI(aq) 
 KI(aq)  H2O(l)
The water could be evaporated to isolate the KI.
(b)
Acids react with carbonates to form carbon dioxide gas. Again, chose the acid and carbonate salt so that
KI is produced.
2HI(aq)  K2CO3(aq) 
 2KI(aq)  CO2(g)  H2O(l)
4.124
The reaction is too violent. This could cause the hydrogen gas produced to ignite, and an explosion could
result.
4.125
All three products are water insoluble. Use this information in formulating your answer.
4.126
(a)
MgCl2(aq)  2NaOH(aq) 
 Mg(OH)2(s)  2NaCl(aq)
(b)
AgNO3(aq)  NaI(aq) 
 AgI(s)  NaNO3(aq)
(c)
3Ba(OH)2(aq)  2H3PO4(aq) 
 Ba3(PO4)2(s)  6H2O(l)
The solid sodium bicarbonate would be the better choice. The hydrogen carbonate ion, HCO 3, behaves as a
Brønsted base to accept a proton from the acid.
HCO3(aq)  H(aq) 
 H2CO3(aq) 
 H2O(l)  CO2(g)
94
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
The heat generated during the reaction of hydrogen carbonate with the acid causes the carbonic acid, H 2CO3,
that was formed to decompose to water and carbon dioxide.
The reaction of the spilled sulfuric acid with sodium hydroxide would produce sodium sulfate, Na2SO4, and
water. There is a possibility that the Na2SO4 could precipitate. Also, the sulfate ion, SO42 is a weak base;
therefore, the “neutralized” solution would actually be basic.
H2SO4(aq)  2NaOH(aq) 
 Na2SO4(aq)  2H2O(l)
Also, NaOH is a caustic substance and therefore is not safe to use in this manner.
4.127
(a)
(b)
(c)
(d)
(e)
4.128
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
A soluble sulfate salt such as sodium sulfate or sulfuric acid could be added. Barium sulfate would
precipitate leaving sodium ions in solution.
Potassium carbonate, phosphate, or sulfide could be added which would precipitate the magnesium
cations, leaving potassium cations in solution.
Add a soluble silver salt such as silver nitrate. AgBr would precipitate, leaving nitrate ions in solution.
Add a solution containing a cation other than ammonium or a Group 1A cation to precipitate the
phosphate ions; the nitrate ions will remain in solution.
Add a solution containing a cation other than ammonium or a Group 1A cation to precipitate the
carbonate ions; the nitrate ions will remain in solution.
Table salt, NaCl, is very soluble in water and is a strong electrolyte. Addition of AgNO 3 will precipitate
AgCl.
Table sugar or sucrose, C12H22O11, is soluble in water and is a nonelectrolyte.
Aqueous acetic acid, CH3COOH, the primary ingredient of vinegar, is a weak electrolyte. It exhibits all
of the properties of acids (Section 4.3).
Baking soda, NaHCO3, is a water-soluble strong electrolyte. It reacts with acid to release CO2 gas.
Addition of Ca(OH)2 results in the precipitation of CaCO3.
Washing soda, Na2CO310H2O, is a water-soluble strong electrolyte. It reacts with acids to release CO 2
gas. Addition of a soluble alkaline-earth salt will precipitate the alkaline-earth carbonate. Aqueous
washing soda is also slightly basic (Section 4.3).
Boric acid, H3BO3, is weak electrolyte and a weak acid.
Epsom salt, MgSO47H2O, is a water-soluble strong electrolyte. Addition of Ba(NO3)2 results in the
precipitation of BaSO4. Addition of hydroxide precipitates Mg(OH)2.
Sodium hydroxide, NaOH, is a strong electrolyte and a strong base. Addition of Ca(NO 3)2 results in the
precipitation of Ca(OH)2.
Ammonia, NH3, is a sharp-odored gas that when dissolved in water is a weak electrolyte and a weak
base. NH3 in the gas phase reacts with HCl gas to produce solid NH4Cl.
Milk of magnesia, Mg(OH)2, is an insoluble, strong base that reacts with acids. The resulting
magnesium salt may be soluble or insoluble.
CaCO3 is an insoluble salt that reacts with acid to release CO2 gas. CaCO3 is discussed in the
Chemistry in Action essays entitled, “An Undesirable Precipitation Reaction” and “Metal from the Sea”
in Chapter 4.
With the exception of NH3 and vinegar, all the compounds in this problem are white solids.
4.129
Reaction 1: SO32(aq)  H2O2(aq) 
 SO42(aq)  H2O(l)
Reaction 2: SO42(aq)  BaCl2(aq) 
 BaSO4(s)  2Cl(aq)
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.130
95
The balanced equation for the reaction is:
XCl(aq)  AgNO3(aq) 
 AgCl(s)  XNO3(aq)
where X  Na, or K
From the amount of AgCl produced, we can calculate the moles of XCl reacted (X  Na, or K).
1.913 g AgCl 
1 mol AgCl
1 mol XCl

 0.01334 mol XCl
143.4 g AgCl 1 mol AgCl
Let x  number of moles NaCl. Then, the number of moles of KCl  0.01334 mol  x. The sum of the NaCl
and KCl masses must equal the mass of the mixture, 0.8870 g. We can write:
mass NaCl  mass KCl  0.8870 g

58.44 g NaCl  
74.55 g KCl 
 x mol NaCl 
  (0.01334  x) mol KCl 
  0.8870 g
1 mol NaCl  
1 mol KCl 

x  6.673  103  moles NaCl
mol KCl  0.01334  x  0.01334 mol  (6.673  103 mol)  6.667  103 mol KCl
Converting moles to grams:
mass NaCl  (6.673  103 mol NaCl) 
mass KCl  (6.667  103 mol KCl) 
58.44 g NaCl
 0.3900 g NaCl
1 mol NaCl
74.55 g KCl
 0.4970 g KCl
1 mol KCl
The percentages by mass for each compound are:
% NaCl 
% KCl 
0.3900 g
 100%  43.97% NaCl
0.8870 g
0.4970 g
 100%  56.03% KCl
0.8870 g
4.131
Cl2O (Cl  1)
Cl2O7 (Cl  7)
Cl2O3 (Cl  3)
ClO2 (Cl  4)
4.132
The number of moles of oxalic acid in 5.00  102 mL is:
Cl2O6 (Cl  6)
0.100 mol H 2C2O4
 (5.00  102 mL)  0.0500 mol H 2C2O4
1000 mL soln
The balanced equation shows a mole ratio of 1 mol Fe2O3 : 6 mol H2C2O4. The mass of rust that can be
removed is:
0.0500 mol H2 C2 O4 
1 mol Fe2 O3
159.7 g Fe2 O3

 1.33 g Fe2O3
6 mol H2 C2 O4
1 mol Fe2 O3
96
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.133
Since aspirin is a monoprotic acid, it will react with NaOH in a 1:1 mole ratio.
First, calculate the moles of aspirin in the tablet.
12.25 mL soln 
0.1466 mol NaOH 1 mol aspirin

 1.796  103 mol aspirin
1000 mL soln
1 mol NaOH
Next, convert from moles of aspirin to grains of aspirin.
1.796  103 mol aspirin 
4.134
The precipitation reaction is:
180.2 g aspirin
1 grain

 4.99 grains aspirin in one tablet
1 mol aspirin
0.0648 g
Ag(aq)  Br(aq) 
 AgBr(s)
In this problem, the relative amounts of NaBr and CaBr2 are not known. However, the total amount of Br in
the mixture can be determined from the amount of AgBr produced. Let’s find the number of moles of Br.
1.6930 g AgBr 
1 mol AgBr
1 mol Br 

 9.015  103 mol Br 
187.8 g AgBr 1 mol AgBr
The amount of Br comes from both NaBr and CaBr2. Let x  number of moles NaBr. Then, the number of
9.015  103 mol  x
. The moles of CaBr2 are divided by 2, because 1 mol of CaBr2
2
produces 2 moles of Br. The sum of the NaBr and CaBr2 masses must equal the mass of the mixture, 0.9157 g.
We can write:
moles of CaBr2 
mass NaBr  mass CaBr2  0.9157 g

199.9 g CaBr2 
102.9 g NaBr   9.015  103  x 
  0.9157 g
 mol CaBr2 
 x mol NaBr 
  

1 mol NaBr  
2
1 mol CaBr2 




2.95x  0.01465
x  4.966  103  moles NaBr
Converting moles to grams:
mass NaBr  (4.966  103 mol NaBr) 
102.9 g NaBr
 0.5110 g NaBr
1 mol NaBr
The percentage by mass of NaBr in the mixture is:
% NaBr 
4.135
(a)
0.5110 g
 100%  55.80% NaBr
0.9157 g
The balanced equations are:
1) Cu(s)  4HNO3(aq) 
 Cu(NO3)2(aq)  2NO2(g)  2H2O(l)
Redox
2) Cu(NO3)2(aq)  2NaOH(aq) 
 Cu(OH)2(s)  2NaNO3(aq)
Precipitation
heat
3) Cu(OH)2(s) 
 CuO(s)  H2O(g)
Decomposition
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
(b)
(c)
4.136
4) CuO(s)  H2SO4(aq) 
 CuSO4(aq)  H2O(l)
Acid-Base
5) CuSO4(aq)  Zn(s) 
 Cu(s)  ZnSO4(aq)
Redox
6) Zn(s)  2HCl(aq) 
 ZnCl2(aq)  H2(g)
Redox
97
1 mol Cu
 1.03 mol Cu . The mole ratio between
63.55 g Cu
product and reactant in each reaction is 1:1. Therefore, the theoretical yield in each reaction is
1.03 moles.
We start with 65.6 g Cu, which is 65.6 g Cu 
1)
1.03 mol 
187.6 g Cu(NO3 )2
 193 g Cu(NO3 )2
1 mol Cu(NO3 )2
2)
1.03 mol 
97.57 g Cu(OH)2
 1.00  102 g Cu(OH)2
1 mol Cu(OH)2
3)
1.03 mol 
79.55 g CuO
 81.9 g CuO
1 mol CuO
4)
1.03 mol 
159.6 g CuSO4
 164 g CuSO4
1 mol CuSO4
5)
1.03 mol 
63.55 g Cu
 65.5 g Cu
1 mol Cu
All of the reaction steps are clean and almost quantitative; therefore, the recovery yield should be high.
There are two moles of Cl per one mole of CaCl2.
(a)
25.3 g CaCl2 
1 mol CaCl2
2 mol Cl

 0.456 mol Cl
111.0 g CaCl2 1 mol CaCl2
Molarity Cl  
(b)
0.456 mol Cl
 1.40 mol/L  1.40 M
0.325 L soln
We need to convert from mol/L to grams in 0.100 L.
1.40 mol Cl 35.45 g Cl

 0.100 L soln  4.96 g Cl 

1 L soln
1 mol Cl
4.137
(a)
CaF2(s)  H2SO4(aq) 
 2HF(g)  CaSO4(s)
2NaCl(s)  H2SO4(aq) 
 2HCl(aq)  Na2SO4(aq)
(b)
HBr and HI cannot be prepared similarly, because Br and I would be oxidized to the element, Br2 and
I2, respectively.
2NaBr(s)  2H2SO4(aq) 
 Br2(l)  SO2(g)  Na2SO4(aq)  2H2O(l)
(c)
PBr3(l)  3H2O(l) 
 3HBr(g)  H3PO3(aq)
98
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.138
(a)
The precipitation reaction is:
Mg2(aq)  2OH(aq) 
 Mg(OH)2(s)
The acid-base reaction is:
Mg(OH)2(s)  2HCl(aq) 
 MgCl2(aq)  2H2O(l)
The redox reactions are:
Mg2  2e 
 Mg
2Cl 
 Cl2  2e
MgCl2 
 Mg  Cl2
4.139
(b)
NaOH is much more expensive than CaO.
(c)
Dolomite has the advantage of being an additional source of magnesium that can also be recovered.
Electric furnace method:
P4(s)  5O2(g) 
 P4O10(s)
redox
P4O10(s)  6H2O(l) 
 4H3PO4(aq)
acid-base
Wet process:
Ca5(PO4)3F(s)  5H2SO4(aq) 
 3H3PO4(aq)  HF(aq)  5CaSO4(s)
This is a precipitation and an acid-base reaction.
4.140
(a)
(b)
NH4(aq)  OH(aq) 
 NH3(aq)  H2O(l)
From the amount of NaOH needed to neutralize the 0.2041 g sample, we can find the amount of the
0.2041 g sample that is NH4NO3.
First, calculate the moles of NaOH.
0.1023 mol NaOH
 24.42 mL soln  2.498  103 mol NaOH
1000 mL of soln
Using the mole ratio from the balanced equation, we can calculate the amount of NH 4NO3 that reacted.
(2.498  103 mol NaOH) 
1 mol NH 4 NO3 80.05 g NH 4 NO3

 0.2000 g NH 4 NO3
1 mol NaOH
1 mol NH 4 NO3
The purity of the NH4NO3 sample is:
% purity 
0.2000 g
 100%  97.99%
0.2041 g
4.141
In a redox reaction, electrons must be transferred between reacting species. In other words, oxidation
numbers must change in a redox reation. In both O2 (molecular oxygen) and O3 (ozone), the oxidation
number of oxygen is zero. This is not a redox reaction.
4.142
Using the rules for assigning oxidation numbers given in Section 4.4, H is 1, F is 1, so the oxidation
number of O must be zero.
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.143

(a)
H3 O
NH4
4.144
H2 O

(b)



OH


H2 O



NH2
99
NH3
NH3
The balanced equation is:
3CH3CH2OH  2K2Cr2O7  8H2SO4 
 3CH3COOH  2Cr2(SO4)3  2K2SO4  11H2O
From the amount of K2Cr2O7 required to react with the blood sample, we can calculate the mass of ethanol
(CH3CH2OH) in the 10.0 g sample of blood.
First, calculate the moles of K2Cr2O7 reacted.
0.07654 mol K 2Cr2O7
 4.23 mL  3.24  104 mol K 2Cr2O7
1000 mL soln
Next, using the mole ratio from the balanced equation, we can calculate the mass of ethanol that reacted.
3.24  104 mol K 2Cr2O7 
3 mol ethanol
46.07 g ethanol

 0.0224 g ethanol
2 mol K 2 Cr2 O7
1 mol ethanol
The percent ethanol by mass is:
% by mass ethanol 
0.0224 g
 100%  0.224%
10.0 g
This is well above the legal limit of 0.1 percent by mass ethanol in the blood. The individual should be
prosecuted for drunk driving.
4.145
Notice that nitrogen is in its highest possible oxidation state (5) in nitric acid. It is reduced as it decomposes
to NO2.
4HNO3 
 4NO2  O2  2H2O
The yellow color of “old” nitric acid is caused by the production of small amounts of NO 2 which is a brown
gas. This process is accelerated by light.
4.146
(a)
Zn(s)  H2SO4(aq) 
 ZnSO4(aq)  H2(g)
(b)
2KClO3(s) 
 2KCl(s)  3O2(g)
(c)
Na2CO3(s)  2HCl(aq) 
 2NaCl(aq)  CO2(g)  H2O(l)
(d)
NH4NO2(s) 
 N2(g)  2H2O(g)
heat
100
CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS
4.147
(a)
The precipitate CaSO4 formed over Ca preventing the Ca from reacting with the sulfuric acid.
(b)
Aluminum is protected by a tenacious oxide layer with the composition Al2O3.
(c)
These metals react more readily with water.
2Na(s)  2H2O(l) 
 2NaOH(aq)  H2(g)
(d)
The metal should be placed below Fe and above H.
(e)
Any metal above Al in the activity series will react with Al3. Metals from Mg to Li will work.
4.148
Because the volume of the solution changes (increases or decreases) when the solid dissolves.
4.149
Place the following metals in the correct positions on the periodic table framework provided in the problem.
(a) Li, Na
(b) Mg, Fe
(c) Zn, Cd
Two metals that do not react with water or acid are Ag and Au.
4.150
4.151
NH4Cl exists as NH4 and Cl. To form NH3 and HCl, a proton (H) is transferred from NH4 to Cl.
Therefore, this is a Brønsted acid-base reaction.
(a)
First Solution:
0.8214 g KMnO4 
M 
1 mol KMnO4
 5.199  103 mol KMnO4
158.0 g KMnO4
5.199  103 mol KMnO4
mol solute

 1.040  102 M
L of soln
0.5000 L
Second Solution:
M1V1  M2V2
(1.040  102 M)(2.000 mL)  M2(1000 mL)
M2  2.080  105 M
Third Solution:
M1V1  M2V2
(2.080  105 M)(10.00 mL)  M2(250.0 mL)
M2  8.320  107 M
(b)
From the molarity and volume of the final solution, we can calculate the moles of KMnO 4. Then, the
mass can be calculated from the moles of KMnO4.
8.320  107 mol KMnO4
 250 mL  2.080  107 mol KMnO4
1000 mL of soln
2.080  107 mol KMnO4 
158.0 g KMnO4
 3.286  105 g KMnO4
1 mol KMnO4
This mass is too small to directly weigh accurately.